Introduction to Finite Fields (cont.)

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1 Chapter 6 Introduction to Finite Fields (cont.) 6.1 Recall Theorem. Z m is a field m is a prime number. Theorem (Subfield Isomorphic to Z p ). Every finite field has the order of a power of a prime number p and contains a subfield isomorphic to Z p. Corollary (Isomorphism to Z p ). Any field F with prime order p is isomorphic to Z p. Theorem (Prime Power Fields). There exists a field F of order n. n is a power of a prime number. 6.2 Polynomials over a field Definition 6.1. Let F be a field. The set n F [x] := { a i x i a i F, n Z 0 } i=0 is called the polynomial ring over F. An element of F [x] is called a polynomial over F. For a polynomial f(x) = n a i x i, if a n 0 then the integer n is called the degree of f(x), i=0 denoted by deg(f(x)), (for convenience, we define deg(0) = ). Furthermore, a nonzero polynomial f(x) = n a i x i of degree n is said to be monic if a n = 1. A polynomial f(x) i=0 59

2 60 Math 422. Coding Theory of positive degree is said to be reducible (over F ) if there exist two polynomials g(x) and h(x) over F such that deg(g(x)) < deg(f(x)), deg(h(x)) < deg(f(x)) and f(x) = g(x)h(x). Otherwise, the polynomial f(x) of positive degree is said to be irreducible (over F ). Example ). The polynomial f(x) = x 4 +2x 6 Z 3 [x] is of degree 6. It is reducible as f(x) = x 4 (1 + 2x 2 ). 2). The polynomial g(x) = 1+x+x 2 Z 2 [x] is of degree 2. It is irreducible. Otherwise, it would have a linear factor x or x + 1; i.e., 0 or 1 would be a root of g(x), but g(0) = g(1) = 1 Z 2. 3). Using the same arguments as in 2), we can show that both 1+x+x 3 and 1+x 2 +x 3 are irreducible over Z 2 as they have no linear factors. Definition 6.2 (Division Rule). Let f(x) F [x] be a polynomial of degree n 1. Then, for any polynomial g(x) F [x], there exists a unique pair (s(x), r(x)) of polynomials with deg(r(x)) < deg(f(x)) or r(x) = 0 such that g(x) = s(x)f(x)+r(x). The polynomial r(x) is called the (principal) remainder of g(x) divided by f(x). If r(x) = 0, then we call f(x) divides g(x) or f(x) is a divisor of g(x) and g(x) is divisible by f(x) g(x) is called a multiple of f(x). Example 6.2. For example, let f(x) = 1 + x 2 and g(x) = x + 2x 4 be two polynomials in Z 5 [x]. Since we have g(x) = x+2x 4 = (3+2x 2 )(1+x 2 )+(2+x) = (3+2x 2 )f(x)+(2+x), the remainder of g(x) divided by f(x) is 2 + x. Lecture 18, March 17, 2011 Lemma 6.3 (Linear Factors). A polynomial f(x) has a linear factor x a. f(a) = 0. Proof. By the division rule, f(x) has a linear factor x a. f(x) = (x a)g(x), i.e., the principal reminder is 0. f(a) = 0. Corollary 6.4 (Irreducible 2nd or 3rd Degree Polynomials). A polynomial f(x) in F [x] of degree 2 or 3 is irreducible. f(a) 0 for all a F. Proof. f(x) can be factored into polynomials of smaller degree. it has at least one linear factor (x a). f(a) = 0, by Lemma. Example ). The polynomial g(x) = 1 + x + x 2 Z 2 [x] is of degree 2. It is irreducible. Since g(0) = g(1) = 1 Z 2.

3 6.2. Polynomials over a field 61 2). Using the corollary, we can show that both 1 + x + x 3 and 1 + x 2 + x 3 are irreducible over Z 2. Analogous to the integral ring Z, we can introduce the following notions. Definition 6.5. Let f(x), g(x) F [x] be two nonzero polynomials. The greatest common divisor of f(x), g(x), denoted by gcd(f(x), g(x)), is the monic polynomial of the highest degree which is a divisor of both f(x) and g(x). In particular, we say that f(x) is coprime (or prime) to g(x) if gcd(f(x), g(x)) = 1. The least common multiple of f(x), g(x), denoted by lcm(f(x), g(x)), is the monic polynomial of the lowest degree which is a multiple of both f(x) and g(x). Remark. 1). If f(x) and g(x) have the following factorizations: f(x) = ap 1 (x) e 1 p 2 (x) e2 p n (x) en, g(x) = bp 1 (x) d1 p n (x) dn, where a, b F \ {0}, e i, d i 0 and p i (x) are distinct monic irreducible polynomials, then gcd(f(x), g(x)) = p 1 (x) min{e 1,d 1} p n (x) min{en,dn} and lcm(f(x), g(x)) = p 1 (x) max{e 1,d 1} p n (x) max{en,dn}. 2). Let f(x), g(x) F [x] be two nonzero polynomials. Then there exist two polynomials u(x), v(x) with deg(u(x)) < deg(g(x)) and deg(v(x)) < deg(f(x)) such that gcd(f(x), g(x)) = u(x)f(x) + v(x)g(x). 3). It is easily shown that gcd(f(x)h(x), g(x)) = gcd(f(x), g(x)) if gcd(h(x), g(x)) = 1. Definition 6.6. Let f(x) be a fixed polynomial in F [x]. Two polynomials g(x) and h(x) in F [x] is said to be congruent modulo f(x), symbolized by g(x) h(x)(mod f(x)) if g(x) h(x) is divisible by f(x). By the division rule, any polynomial g(x) in F [x] is congruent module f(x) to a unique polynomial r(x) of degree less than deg(f(x)) (r(x) is just the principal remainder when g(x) is divided by f(x)). We denote F [x]/(f(x)) or F [x]/f(x) the set of polynomials in F [x] of degree less than deg(f(x)), with addition and multiplication carried out modulo f(x) as follows:

4 62 Math 422. Coding Theory Suppose g(x) and h(x) belong to F [x]/f(x). Then the sum g(x) + h(x) in F [x]/f(x) is the same as the sum in F [x], because deg(g(x) + h(x)) < deg(f(x)). The product g(x)h(x) is the principal remainder when g(x)h(x) is divided by f(x). There are many analogies between the integral ring Z and a polynomial ring F [x]. Theorem 6.1. Let f(x) be a polynomial over a field F of degree 1. Then F [x]/(f(x)), together with the addition and multiplication, forms a ring. Furthermore, F [x]/(f(x)) is a field if and only if f(x) is irreducible. Proof. It is easy to verify that F [x]/(f(x)) is a ring. By applying exactly the same arguments as in the proof of Theorem 1.5, we can prove the second part. Example ). Consider the ring R[x]/(1 + x 2 ) = {a + bx a, b R}. It is a field since 1 + x 2 is irreducible over R. In fact, it is isomorphic to the complex field C! To see this, we just replace x in R[x]/(1 + x 2 ) by the imaginary unit i. 2). Consider the ring Z 2 [x]/(1 + x 2 ) = {0, 1, x, 1 + x}. We construct the addition and multiplication tables as follows. We see from the multiplication table that Z 2 [x]/(1 + x 2 ) is not a field as (1 + x)(1 + x) = 0. 3). Consider the ring Z 2 [x]/(1 + x + x 2 ) = {0, 1, x, 1 + x}. As 1 + x + x 2 is irreducible over Z 2, the ring Z 2 [x]/(1 + x + x 2 ) is in fact a field. This can also be verified by the addition and multiplication tables. Remark. If F = F q is a finite field, then for polynomial f(x) with degree n we have F [x]/(f(x)) = q n. Indeed, the ring F [x]/f(x) consists of all polynomials of degree n 1. Each of the n coefficients of such a polynomial belongs to F q, i.e., has q choices. 6.3 Structure of finite fields We list some facts about finite fields in this section: Lemma 6.7. For every element α of a finite field F q, we have α q = α. Corollary 6.8. Let F be a subfield of E with F = q. Then an element α of E lies in F if and only if α q = α. Theorem 6.2. For any prime p and integer n 1, there exists a unique finite field of p n elements.

5 6.3. Structure of finite fields 63 Lecture 19, March 22, 2011 Recall. For an irreducible polynomial f(x) of degree n over a field F, let α be a root of f(x) = 0. Then the field F [x]/(f(x)) can be represented as F [α] = {a 0 + a 1 α + + a n 1 α n 1 a i F } if we replace x in F [x]/(f(x)) by α. An advantage of using F [α] to replace the field F [x]/(f(x)) is that we can avoid the confusion between an element of F [x]/(f(x)) and a polynomial over F. Definition 6.9 (Primitive element). An element α in a finite field F q primitive element (or generator) of F q if F q = {0, α, α 2,..., α q 1 }. is called a Example 6.5. Consider the field F 4 = F 2 [x]/(x 2 +x+1) = F 2 [α] = {0, 1, x, x+1}, where α is a root of the irreducible polynomial 1 + x + x 2 F 2 [x]. Then we have α 2 = (1 + α) = 1 + α, α 3 = α(α 2 ) = α(1 + α) = α + α 2 = 1. Thus, F 4 = {0, α, α 2, α 3 }, so α is a primitive element. Definition 6.10 (Order). The order of a nonzero element α F q, denoted by ord(α), is the smallest positive integer k such that α k = 1. Example 6.6. Since there are no linear factors for the polynomial 1+x 2 over F 3, 1+x 2 is irreducible over F 3. Consider the element α in the field F 9 = F 3 [x]/(x 2 +1) = F 3 [α], where α is a root of 1 = x 2. Then α 2 = 1, α 3 = α(α 2 ) = α and α 4 = (α 2 ) 2 = ( 1) 2 = 1. This means that ord(α) = 4. Lemma ). The order ord(α) divides q 1 for every α F q. 2). For two nonzero elements α, β F q, if gcd(ord(α), ord(β)) = 1, then ord(α β) = ord(α) ord(β). Theorem ). A nonzero element of F q is a primitive element if and only if its order is q 1. 2). Every finite field has at least one primitive element. Remark. 1). Primitive elements are not unique. 2). If α is a root of an irreducible polynomial f(x) of degree m over F q, and it is also a primitive element of F q m = F [x]/f(x) = F q [α], then every element in F q m can be represented both as a polynomial in α and as a power of α, since F q m = {a 0 + a 1 α + + a m 1 α m 1 a i F q } = {0, α, α 2,..., α qm 1 }.

6 64 Math 422. Coding Theory Addition for the elements of F q m is easily carried out if the elements are represented as polynomials in α, whilst multiplication is easily done if the elements are represented as powers of α. Example ). Let α be a root of 1 + x + x 3 F 2 [x]. Hence, F 8 = F 2 [α]. The order of α is a divisor of 8 1 = 7. Thus, ord(α) = 7 and α is a primitive element. In fact, any nonzero element in F 8 except 1 is a primitive element, since all the elements like this is not order of 1. 2). Let α be a root of 1 + 2x + x 3 F 3 [x]. This polynomial is irreducible over F 3 as it has no linear factors. Hence, F 27 = F 3 [α]. The order of α is a divisor of 27 1 = 26. Thus, ord(α) is 2, 13 or 26. First, ord(α) 2; otherwise, α would be 1 or 1, neither of which is a root of 1 + 2x + x 3. Furthermore, we have α 13 = 1 1, indeed α 13 = α α 3 (α 3 ) 3 = α ( 2α 1) ( 2α 1) 3 = α ( 2α 1) ( 8α 3 1) = α ( 2α 1) (α 3 1) = α (α 1) (α 2) = α 3 + 2α = 1 Thus, ord(α) = 26 and α is a primitive element of F Minimal polynomials Definition 6.12 (Minimal polynomial). A minimal polynomial of an element α F q m with respect to F q is a nonzero monic polynomial f(x) of the least degree in F q [x] such that f(α) = 0. Example 6.8. Let α be a root of the polynomial 1 + x + x 2 F 2 [x]. It is clear that the two linear polynomials x and 1 + x are not minimal polynomials of α. Therefore, 1 + x + x 2 is a minimal polynomial of α. Since 1 + (1 + α) + (1 + α) 2 = α α 2 = 1 + α + α 2 = 0 and 1 + α is not a root of x or 1 + x, thus 1 + x + x 2 is also a minimal polynomial of 1 + α. Theorem 6.4. Let β F p r. If f(x) F p [x] has β as a root, then f(x) is divisible by the minimal polynomial of β. Corollary The minimal polynomial of an element of a field F q divides x q x. Theorem ). The minimal polynomial of an element of F q m with respect to F q exists and is unique. It is also irreducible over F q.

7 6.4. Minimal polynomials 65 2). If a monic irreducible polynomial M(x) F q [x] has α F q m as a root, then it is the minimal polynomial of α with respect to F q. Definition 6.14 (cyclotomic coset). Let r be co-prime to q. The cyclotomic coset of q (or q-cyclotomic coset) modulo r containing i is defined by C i = {(i q j (mod r)) Z r j = 0, 1,...}. A subset {i 1,..., i t } of Z r is called a complete set of representatives of cyclotomic cosets of q modulo r if C i1,..., C it are distinct and t j=1 C i j = Z r. Remark. 1). It is easy to verify that two cyclotomic cosets are either equal or disjoint. Hence, the cyclotomic cosets partition Z r. Lecture 20, March 24, ). If r = q m 1 for some m 1, each cyclotomic coset contains at most m elements, as q m 1(mod q m 1). 3). It is easy to see that, in the case of r = q m 1 for some m 1, if gcd(i, q m 1) = 1 then C i = m. Example ). Consider the cyclotomic cosets of 2 modulo 15: C 0 = {0}, C 1 = {1, 2, 4, 8}, C 3 = {3, 6, 9, 12}, C 5 = {5, 10}, C 7 = {7, 11, 13, 14}. Thus, C 1 = C 2 = C 4 = C 8, and so on. The set {0, 1, 3, 5, 7} is a complete set of representatives of cyclotomic cosets of 2 modulo 15. The set {0, 1, 6, 10, 7} is also a complete set of representatives of cyclotomic cosets of 2 modulo 15. 2). Consider the cyclotomic cosets of 3 modulo 26: C 0 = {0}, C 1 = {1, 3, 9}, C 2 = {2, 6, 18}, C 4 = {4, 12, 10}, C 5 = {5, 15, 19}, C 7 = {7, 21, 11}, C 8 = {8, 24, 20}, C 13 = {13}, C 14 = {14, 16, 22}, C 17 = {17, 25, 23}. In this case, we have C 1 = C 3 = C 9, and so on. The set {0, 1, 2, 4, 5, 7, 8, 13, 14, 17} is a complete set of representatives of cyclotomic cosets of 3 modulo 26. field. We are now ready to determine the minimal polynomials for all the elements in a finite

8 66 Math 422. Coding Theory Theorem 6.6. Let α be a primitive element of F q m. Then the minimal polynomial of α i is M (i) (x) = (x α j ), j C i where C i is the unique cyclotomic coset of q modulo q m 1 containing i. Remark. 1). The degree of the minimal polynomial of α i is equal to the size of the cyclotomic coset containing i. 2). From the Theorem, we know that α i and α k have the same minimal polynomial if and only if i, k are in the same cyclotomic coset. Example Let α be a root of 2 + x + x 2 F 3 [x]; i.e., 2 + α + α 2 = 0. Then the minimal polynomial of α as well as α 3 is 2 + x + x 2. The minimal polynomial of α 2 is M (2) (x) = (x α j ) = (x α 2 )(x α 6 ) = α 8 (α 2 + α 6 )x + x 2. j C 2 We know that α 8 = 1 as α F 9. To find M (2) (x), we have to simplify α 2 + α 6. We make use of the relationship above to obtain α 2 + α 6 = (1 α) + (1 α) 3 = 2 α α 3 = 2 α α(1 α) = 2 2α + α 2 = 0. Hence, the minimal polynomial of α 2 is 1 + x 2. In the same way, we may obtain the minimal polynomial 2 + 2x + x 2 of α 5. The following result will be useful when we study cyclic codes. Theorem 6.7. Let n be a positive integer with gcd(q, n) = 1. Suppose that m is a positive integer satisfying n (q m 1). Let α be a primitive element of F q m and let M (j) (x) be the minimal polynomial of α j with respect to F q. Let {s 1, s 2,..., s t } be a complete set of representatives of cyclotomic cosets of q modulo n. Then the polynomial x n 1 has the factorization into monic irreducible polynomials over F q : x n 1 = t i=1 M ( (qm 1)s i n ) (x). Remark. 1). In the theorem, the set {s 1, s 2,..., s t } is a complete set of representatives of cyclotomic cosets of q modulo n, but by Theorem 6.6 the minimal polynomial M ( (qm 1)s i n ) (x) is given through complete set of representatives of cyclotomic cosets of q modulo q m 1.

9 6.4. Minimal polynomials 67 2). The factorization of x n 1 into monic irreducible polynomials over F q doesn t depend on the choice of m. Corollary Let n be a positive integer with gcd(q, n) = 1. Then the number of monic irreducible factors of x n 1 over F q is equal to the number of cyclotomic cosets of q modulo n. Example ). Consider the polynomial x 13 1 over F 3. It is easy to check that {0, 1, 2, 4, 7} is a complete set of representatives of cyclotomic cosets of 3 modulo 13. Since 13 is a divisor of 3 3 1, we consider the field F 27. Let α be a root of 1 + 2x + x 3. By the Example, α is a primitive element of F 27. By Example, we know all the cyclotomic cosets of 3 modulo 26 containing multiples of 2. Hence, we obtain M (0) (x) = x + 2, M (2) (x) = j C 2 (x α j ) = (x α 2 )(x α 6 )(x α 18 ) = 2 + x + x 2 + x 3, M (4) (x) = j C 4 (x α j ) = (x α 4 )(x α 12 )(x α 10 ) = 2 + x 2 + x 3 M (8) (x) = j C 8 (x α j ) = (x α 8 )(x α 20 )(x α 24 ) = 2 + 2x + 2x 2 + x 3 M (14) (x) = j C 14 (x α j ) = (x α 14 )(x α 16 )(x α 22 ) = 2 + 2x + x 3. By the Theorem, we obtain the factorization of x 13 1 over F 3 into monic irreducible polynomials: x 13 1 = M (0) (x) M (2) (x) M (4) (x) M (8) (x) M (14) (x) = (x + 2)(2 + x + x 2 + x 3 )(2 + x 2 + x 3 )(2 + 2x + 2x 2 + x 3 )(2 + 2x + x 3 ). 2). We can also discuss the polynomial x 21 1 over F 2. It is easy to check that {0, 1, 3, 5, 7, 9} is a complete set of representatives of cyclotomic cosets of 2 modulo 21. Since 21 is a divisor of 2 6 1, we consider the field F 64. Let α be a root of 1 + x + x 6. It can be verified that α is a primitive element of F 64 (check that α 3 1, α 7 1, α 9 1 and α 21 1). We list the cyclotomic cosets of 2 modulo 63 containing multiples of 3: C 0 = {0}, C 3 = {36, 12, 24, 48, 33}, C 9 = {9, 18, 36}, C 15 = {15, 30, 60, 57, 51, 39}, C 21 = {21, 42}, C 27 = {27, 54, 45}.

10 68 Math 422. Coding Theory Hence, we obtain M (0) (x) = x + 1, M (3) (x) = j C 3 (x α j ) = 1 + x + x 2 + x 4 + x 6, M (9) (x) = j C 9 (x α j ) = 1 + x 2 + x 3, M (15) (x) = j C 15 (x α j ) = 1 + x 2 + x 4 + x 5 + x 6, M (21) (x) = j C 21 (x α j ) = 1 + x + x 2, M (27) (x) = j C 27 (x α j ) = 1 + x + x 3. By the above Theorem, we obtain the factorization of x 21 1 over F 2 into monic irreducible polynomials: x 21 1 = M (0) (x) M (3) (x) M (9) (x) M (15) (x) M (21) (x) M (27) (x) = (x + 1) (1 + x + x 2 + x 4 + x 6 ) (1 + x 2 + x 3 ) (1 + x 2 + x 4 + x 5 + x 6 ) (1 + x + x 2 ) (1 + x + x 3 ). Remark. We can find the factorization of x n 1 over F q step by step as following: Step 1: Give a complete set of representatives of cyclotomic cosets of q modulo n. Step 2: Find a positive integer m such that n q m 1. Step 3: Give a primitive element α of the field F q m. Step 4: List the corresponding cyclotomic cosets of q modulo q m 1 and using Theorem 6.6 to write down the minimal polynomials. Step 5: Get the factorization through the minimal polynomials.

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